Below is my code for a 4th order Runge-Kutta but I keep getting an error and I don't see what I am doing wrong. I was hoping somebody could help point me in the right direction to fix my code.

 Akzeptierte Antwort

James Tursa
James Tursa am 20 Apr. 2016
Bearbeitet: James Tursa am 20 Apr. 2016

0 Stimmen

You never set the z2 elements inside your loop, so on the second iteration when z2(i-1) is used (with i=3) z2(2) does not exist. I am guessing you need to add a line like this after setting z1(i):
z2(i) = z2(i-1)+((1/6)*(k1z2+2*k2z2+2*k3z2+k4z2));
Also, it appears you have a typo in your z1(i) assignment. That last * should be a + instead. E.g.,
z1(i) = z1(i-1)+((1/6)*(k1z1+2*k2z1+2*k3z1*k4z1));
should be
z1(i) = z1(i-1)+((1/6)*(k1z1+2*k2z1+2*k3z1+k4z1));
Finally, as a readability advice I would suggest you put some spaces in your code. E.g., I think this is much more readable (and easier to spot bugs):
z1(i) = z1(i-1) + (1/6) * (k1z1 + 2*k2z1 + 2*k3z1 + k4z1);
z2(i) = z2(i-1) + (1/6) * (k1z2 + 2*k2z2 + 2*k3z2 + k4z2);

1 Kommentar

WhatIsMatlab-
WhatIsMatlab- am 20 Apr. 2016
Thank you!! That did the trick. I took your advice on adding the spacing. You are 100% right it makes it a lot easier to see what is happening in my code. Thank you again.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Torsten
Torsten am 20 Apr. 2016

0 Stimmen

dz2dx = @(z1,z2) 2*z2 5*z1;
There is no connector between z2 and 5 in the expression above.
Best wishes
Torsten.

1 Kommentar

WhatIsMatlab-
WhatIsMatlab- am 20 Apr. 2016
Sorry there should be a plus sign there. But that doesn't fix the problem. That was an error from copying the code over.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Hilfe-Center und File Exchange

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by